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I wish to prove that $G/K(G)$ is commutative, where $G$ is a group and $K(G)$ denotes the set $$\{aba^{-1}b^{-1}|a,b \in G\}.$$

I tried doing this using the definition (I took $X,Y$ in $K(G)$ and tried proving $XY-YX$ is in $K(G)$ but I failed...)

Edit: as pointed out by @Dylan Moreland, I meant to ask about the subgroup generated by the commutators $\{aba^{-1}b^{-1}|a,b \in G\}$.

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If $G$ is a group, then the expression $XY-YX$ does not make sense at all. –  Henning Makholm Mar 13 '12 at 17:17
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You really want $K(G)$ to be the subgroup generated by the commutators $aba^{-1}b^{-1}$. The set of commutators is not always a subgroup. The easiest example I can think of is the free group on four generators. Also, you should prove that $K(G)$ is a normal (even better, it's characteristic) subgroup of $G$. –  Dylan Moreland Mar 13 '12 at 17:17
    
Make sure you understand what is happening here: when you take the quotient $\Gamma=G/[G,G]$ you're introducing the relations $aba^{-1}b^{-1}=1$ in $\Gamma$, hence $\Gamma$ is commutative. –  lhf Mar 15 '12 at 11:32

3 Answers 3

up vote 9 down vote accepted

You can't, because the set of elements of the form $aba^{-1}b^{-1}$ may fail to be a subgroup; if it is not a subgroup, then you cannot take the quotient modulo $K(G)$, so you cannot even talk about $G/K(G)$.

(And $XY-YX$ makes no sense in groups; that's the commutator of two elements in a ring, not in a group)

(For examples of groups $G$ in which $K(G)$ is not a subgroup, see here, here, and here)

However, if you define $[G,G]$ to be the subgroup generated by the set $K(G)$, then it is easy to show that this subgroup is normal: if $aba^{-1}b^{-1}$ is a generator, then for every $g\in G$, $$g^{-1}(aba^{-1}b^{-1})g = (gag^{-1})(gbg^{-1})(ga^{-1}g^{-1})(gb^{-1}g^{-1})\in K(G),$$ so the group must be normal.

Thus, we can now meaningfully talk about $G/[G,G]$.

To show that $G/[G,G]$ is abelian, note that if $a[G,G]$ and $b[G,G]$ are elements of the quotient, then $$(a[G,G])(b[G,G]) = ab[G,G] = ab(b^{-1}a^{-1}ba)[G,G] = ba[G,G] = (b[G,G])(a[G,G]),$$ with the middle equality because $b^{-1}a^{-1}ba\in [G,G]$. Thus, $G/[G,G]$ is abelian.

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Well, the answer has already been spelt out.

First up, I'd suggest you actually prove that the subgroup generated by $K(G)^\dagger$ is actually a normal subgroup. (Otherwise, you'd be attempting to prove a meaningless statement.)

I'll further ask you to prove a more general statement.

Let $G$ be a group and $K$ be a normal subgroup of $G$. Prove that $G/K$ is abelian if and only if $K \supseteq \langle K(G) \rangle$ where $K(G) $ follows your definition.

Note that the result you're after follows immediately if you take $K=\langle K(G) \rangle$


$\dagger$ The subgroup generated by $K(G)$ is called the Commutator subgroup of $G$ and is a very interesting and important subgroup of $G$.

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As has already been mentioned, you only get a group if you mod out by a normal subgroup. $K(G)$ as you define it is not even a group, let alone a normal subgroup, so you need to talk about the subgroup generated by elements of the form $\{aba^{-1}b^{-1} \mid a, b \in G\}$.

When you mod out by $K(G)$, you're saying that all elements of $K(G)$ are now identified as the identity in $G / K(G)$, so it makes sense intuitively that $G / K(G)$ is commutative. But, it's also easy to prove. For all $a K(G), b K(G) \in G / K(G)$, $$aba^{-1}b^{-1} K(G) = K(G),$$ so that $$ab K(G) = ba K(G)$$ which is equivalent to $$a K(G) \cdot b K(g) = b K(G) \cdot a K(G)$$

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